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March 31, 2015

March 31, 2015

Posted by **Mike** on Sunday, January 6, 2008 at 1:36pm.

I need help. This is for my midterm review, if anyone is willing to help, I'd appreciate it.

a) sin theta= 3/5 and theta is in quadrant 1.

b) tan theta= -2 and theta is in quadrant II

c) sec theta = 2 radical 3/3 and theta is in quadrant II

d) csc theta= -2/3 and theta is in quadrant III.

I need to find all six trigonometric functions for each one ( sin, cos, tan, csc, sec, cot)

- Trigonometric Functions -
**bobpursley**, Sunday, January 6, 2008 at 1:45pmWith each function given, you have two sides of a right triangle. Using the pyth theorm, you can find the other side, and thus, the other functions. We will be happy to check your work.

For example, c) If secTheta = 2 sqrt (1/3), then the sides of the triangle are

opposite figure out

adjacent sqrt 3

hypo= 2

and the opposite will be

4=3+opposite^2 or opposite is +-1, Now in quadrant two, that means sec is negative, so it should have been -2sqrt(1/3), and the opposite (vertical) is 1. From that you get the six functions.

- Trigonometric Functions -
**Damon**, Sunday, January 6, 2008 at 1:52pmI will do c. You try the rest.

Oh no I won't --> sign of sec is negative in quad 2. I think typo

I will do d.

d) csc T = 1/sin T = -2/3

Hey! what is going on here?

the absolute value of sin T may not be greater than 1. It can not be -3/2

Either we are looking at typos or someone is playing games with you.

I will do b I guess

b) draw triangle in quad 2

-1 along -x axis

+2 up at x = -2

now hypotenuse = sqrt (-1^2 + 2^2) = sqrt 5

so

sin = 2/sqrt 5

cos = -1/sqrt 5 = -(1/5)sqrt 5

tan = -2/1

csc = 1/sin = (1/2)sqrt 5

sec = 1/cos = -sqrt 5

cot = 1/tan = -1/2

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